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I want to compute following radical

$$\sqrt[4]{28+16 \sqrt 3}$$

For that, I first tried to rewrite this in terms of exponential.

$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$

We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$

$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$

However, I'm stuck at this step. Could you assist me?

Regards

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  • $\begingroup$ Is there an exponential approach to this question? $\endgroup$
    – Melz
    Dec 22, 2018 at 10:08
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    $\begingroup$ The statment $28 = 2 \cdot 7^{\frac{1}{2}}$ is false. It should be $28=2^2\cdot 7$ or $28^{\frac12} = 2\cdot 7^{\frac{1}{2}}$. $\endgroup$
    – Somos
    Dec 22, 2018 at 10:12
  • $\begingroup$ Read about completing a square $\endgroup$
    – farruhota
    Dec 30, 2018 at 2:21

7 Answers 7

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$$\sqrt[4] {28+16\sqrt 3}=\sqrt[4] {(\sqrt {12})^2 +(\sqrt {16})^2 +2\sqrt {16\cdot 12}}=\sqrt {4+2\sqrt 3}=\sqrt {(\sqrt 3)^2 +(1)^2 +2\sqrt {3\cdot 1}}=\sqrt 3 +1$$

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  • $\begingroup$ Is it a rule or something? $\endgroup$
    – Melz
    Dec 22, 2018 at 10:05
  • $\begingroup$ @Enzo Since you are taking the fourth root, you would like to express $28+16\sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16\sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$ $\endgroup$ Dec 22, 2018 at 10:11
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Hint.

  • $28+16\sqrt3=12+16+2\cdot2\sqrt3\cdot4=(2\sqrt3)^2+4^2+2\cdot2\sqrt3\cdot4=(4+2\sqrt3)^2$

  • $4+2\sqrt3=3+1+2\cdot1\cdot\sqrt3=(\sqrt3)^2+1^2+2\cdot1\cdot\sqrt3=(\sqrt3+1)^2$

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Hint:

Try to write $$28+16 \sqrt 3=(a+b\sqrt 3)^4$$ for suitable $a$ and $b$.

You obtain, reordering the terms \begin{align} (a+b\sqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)\sqrt 3.\\ \end{align} Can you choose $a$ and $b$ so that $$a^4+18a^2b^2+9b^4=28,\quad ab(a^2+3b^2)=4?$$

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To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.

First time round, we can find that $\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$. Subtracting out the norm gives $24 + 16 \sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$, we get $\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$.

Second time round, we can find that $\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2 -8} = \sqrt{4} = 2$. Subtracting out the norm gives $2 + 2\sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{4} = 2$, you get $1 + \sqrt{3}$.

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Note

$$\sqrt[4]{28+16 \sqrt 3}=\sqrt[4]{4(7+2 \sqrt {3\cdot4})} = \sqrt[4]{4(\sqrt4+\sqrt {3})^2}\\ = \sqrt{4+2\sqrt{1\cdot3}}=\sqrt{(1+\sqrt3)^2}=1+\sqrt3 $$

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In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+b\sqrt 3$ for integers $a$ and $b$, then you can solve it like this:

Step 1

($a+b\sqrt 3)^4=28+16\sqrt 3$. So suppose ($a+b\sqrt 3)^2=c+d\sqrt 3$. Then $(c+d\sqrt 3)^2=28+16\sqrt 3$, i.e.

$$(c^2+3d^2)+2cd\sqrt 3=28+16\sqrt 3$$

$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+\dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:

$$c^4-28c^2+192=0$$

This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=\pm4$, and so $d=\pm2$. So we have two possible solutions: $\pm(4+2\sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16\sqrt 3$). Pick the positive solution $4+2\sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).

Step 2

Now you need to solve $(a+b\sqrt 3)^2=4+2\sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.

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Alternatively, you can use Michael Rozenberg's identities, where this identity applies to the question:

$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$

We have $a = 28$, $b = 16^2 \cdot 3 = 768$. Applying this identity gives:

$$\sqrt{28+\sqrt{768}}=\sqrt{\frac{28+\sqrt{16}}{2}}+\sqrt{\frac{28-\sqrt{16}}{2}}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}.$$

Applying the identity once more gives, where $a=4, b=2^2 \cdot 3 = 12$ gives: $$\sqrt{4+\sqrt{12}}=\sqrt{\frac{4+\sqrt{4^2-12}}{2}}+\sqrt{\frac{4-\sqrt{4^2-12}}{2}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1.$$

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